A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm) which has a net charge of –4.0 nC. Determine the resulting charge density on the inner surface of the conducting sphere.

Explanation:

The given data is as follows.

             q = 6.0 nC =

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            .......... (1)

Since, area of the sphere is as follows.

               A = ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

                   =            

                   =

or,               = 4.77

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 .

pe = weightxheight = 2000x1000=2,000,000joules